General solution of integration by parts of int(x^n*e^x)

brandy
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Homework Statement


i have to create a general formula for integral of (x^n * e^x) dx
using whatever method i deem appropriate. (the only way i could think of is by parts)


Homework Equations


int(x^n * e^x)dx
int(uv')dx=uv-int(vu')dx


The Attempt at a Solution


i used integration by parts. so. I am having trouble with the uv part.
so far I've got
n!*e^x * (U) - int(e^x*n!)
U=? something that sums up u-u'-u''-u'''... until x is to the power of 1.

i figured out the function n-(n-1)-(n-2) etc which is = n-n(n-1)/2
i think if i can manipulate it enough it can give me the solution. but idk how

really, i just need a push in the right direction. or some clues or hints or something. ps make it simple, i take a while to understand other peoples working.
 
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ooooh poooooo!
i just realized this doesn't factor in the fact that you end up minusing a new function with a negative in it and so on so the solution ends up as +term -term + term - term etc


also, something that just further comfused me was this:
i read the next part of the question which says that i need to derive this formula which is the solution to the problem


In = xn ex - n*In-1
where In \intxn ex dx
 
brandy said:
In = xn ex - n*In-1
where In \intxn ex dx

Just apply integration by parts once to In and that result would easily follow.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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